# Will the set of algebraic polynomials be dense in space $L_{\infty}(0,1)$?

I know that the set of algebraic polynomials is dense in space $$L_{p}(a,b)$$, where $$1 \leq p < \infty$$ and $$a,b \in \mathbb{R}$$. However, what about $$L_{\infty}(0,1)$$? Will the set of algebraic polynomials be dense in space $$L_{\infty}(0,1)$$?

In my opinion it is not, but is there any example that prove this fact?

• Since $L_{\infty}(0,1)$ is not separable, it cannot have the set of real polynomials as a dense subset. If it were true, the set of algebraic polynomials with rational coefficients would be dense in $L_{\infty}(0,1)$. But that is not possible since the set of rational polynomials is countable. – Darth Lubinus Jun 3 at 18:48
• @DarthLubinus, Could you explain me why this is impossible? – Sh VavilenT Jun 3 at 19:52
• Sure, I'll give some details in an answer! – Darth Lubinus Jun 3 at 19:55

I'm going to try and ellaborate on my previous comment.

First, let's call $$\mathbb{R}[X]$$ the ring of real polynomials, and suppose $$\mathbb{R}[X]$$ was a dense subset of $$L_{\infty}(0,1)$$. Therefore, we can prove that $$\mathbb{Q}[X]$$ (the subset of all rational polynomials) is dense in $$L_{\infty}(0,1)$$ by seeing that it is dense in $$\mathbb{R}[X]$$.

Let $$p(x)=a_{n}x^{n}+...+a_{1}x+a_{0}$$ be a polynomial with coefficients in $$\mathbb{R}$$, and $$\varepsilon>0$$. Then, for every $$a_{i}$$ we can find a $$b_{i}\in \mathbb{Q}$$ such that $$|b_{i}-a_{i}|<\varepsilon/n$$. By taking $$q(x)=b_{n}x^{n}+...+b_{1}x+b_{0}$$, we have that $$q\in \mathbb{Q}[X]$$, and for every $$x\in [0,1]$$,

$$|p(x)-q(x)|=\Big|\sum_{i=0}^{n}(a_{i}-b_{i})x\Big|\leq \sum_{i=0}^{n}|a_{i}-b_{i}||x|\leq n\frac{\varepsilon}{n}=\varepsilon,$$

so $$||p-q||_{\infty}\leq\varepsilon$$. Therefore, $$\mathbb{Q}[X]$$ is dense in $$\mathbb{R}[X]$$ and, using our assumption that $$\mathbb{R}[X]$$ is dense in $$L_{\infty}(0,1)$$, we conclude that $$\mathbb{Q}[X]$$ is dense in $$L_{\infty}(0,1)$$. In particular, $$L_{\infty}(0,1)$$ admits a countable dense subspace, which means that it is separable.

To find our desired contradiction, we need to prove that $$L_{\infty}(0,1)$$ cannot be separable.

Consider the family $$\{f_{t}\colon t\in (0,1)\}\subseteq L_{\infty}(0,1)$$, where $$f_{t}(s)=1$$ if $$s\leq t$$, and $$f_{t}(s)=0$$ otherwise. It is easy to see that, whenever $$t\neq s$$, we have $$||f_{t}-f_{s}||_{\infty}=1$$. Therefore, we can find an uncountable family of disjoint open balls $$A_{t}=B(f_{t},1/3)$$, where $$0 < t < 1$$. Since $$\mathbb{Q}[X]$$ is dense in $$L_{\infty}(0,1)$$, each $$A_{t}$$ must contain at least one polynomial $$q_{t}\in \mathbb{Q}[X]$$, but this is impossible, because it would imply that there is an uncountable amount of rational polynomials! Therefore, $$\mathbb{R}[X]$$ cannot be dense.

In essence, what is happening is that $$L_{\infty}$$ is too big of a space to be the closure of something as comparably small as a countable subset. When $$p<\infty$$, this doesn't happen, and the fact that polynomials are dense in $$L_{p}(0,1)$$ proves that this last space is separable.

Just for the sake of adding another point of view, here is a more "topological" proof (not too different from the one I just did). In metric spaces, separability is the same thing as second countability (the property of having a countable basis for the topology of the metric space). If $$L_{\infty}(0,1)$$ were separable/secound countable, then the family $$B=\{f_{t}\}_{t\in (0,1)}$$ would be another second countable space (as a subspace of one). But $$||f_{t}-f_{s}||_{\infty}=1$$ when $$t\neq s$$, so $$B$$ must be discrete. In that case, $$B$$ must be countable, which is again a contradiction.

I hope this helps!